The Fencing Problem
For the first part of my investigation, I will have to some possible figures for the plot of land, making sure that the perimeter is 1000m.
For my first examination, I will start with quadrilaterals.
Side 1: 100m
Side 2: 400m
Side 1: 200m
Side 2: 300m
Side 1: 250m
Side 2: 250m
Side 1: 450m
Side2: 50m
Side 1: 350m
Side2: 150m
Side 1
Side 2
Area
00 m
400 m
Side*side=
40000 m2
250 m
250 m
Side*side=
62500 m2
300 m
200 m
Side*side=
60000 m2
350 m
50 m
Side*side=
52500 m2
450 m
50 m
Side*side=
22500 m2
The graph and the table indicate that when the width and the height of a regular quadrilateral have the least difference, the Area is at its greatest. This statement is proven by the square. A square has all the sides the same, so the difference is at the smallest since it is at zero. The Area of the square is the greatest, and this leads me to believe that regular shapes- i.e. shapes that have equal sides, have the biggest Area. I will test this prediction out on triangles. I will start with the most regular triangle - the equilateral, and then test a number of isosceles triangles. If my prediction is true then the equilateral triangle will have the biggest Area.
Side 1, 2,3= 333.333
Sin (60)= x/333.33
X=288.66 (height)
Area= ( X base X height
=48112.52m2
Side 2, 3=450m
Base= 100m
150m2 +x2= 350m2
x=316.23m (height)
Area= ( X base X height
= 47494.16m2
Side 2, 3=400m
Base= 200m
00m2 +x2=400m2
x=382.30 (height)
Area=( X base X height
=38729.83m2
Side 2, 3=350m
Base= 300m
50m2+x2=350m2
x=316.23m
Area=( X base X height
=47434.16m2
Side 2, 3=300m
Base=400m
200m2+x2=300m2
x=223.61m
Area=( X base X height
=44721.36m2
Base
Equal sides
Area
00m
450m
22360.68m2
200m
400m
38729.83 m2
300m
350m
47434.16 m2
333(m
333(m
48112.52 m2
400m
300m
44721.36m2
This graph proves my initial theory. The equilateral triangle, which had the least difference between the sides, has the largest Area. The equilateral triangle, which is the most regular of all the triangles since it has equal sides, has the biggest Area. This is similar to the square, which is the most regular of all quadrilaterals. Therefore, I can now use only regular polygons (polygons with equal sides) to gain the biggest Areas. However, I have also made another observation. The three-sided triangle has a smaller Area than the four-sided square. It may be possible that Area increases with the number of sides. Consequently, this is my next hypothesis. To test this new theory, I will have to work out the Area for other regular polygons, with a greater number of sides. I will start off with a pentagon.
- Each side= 200m (1ooo/5)
- All corresponding angles are the same
- Exterior angle = 360o/n or 360/5 = 72o
- Interior angles= 180-72=1080 (angles on a straight line)
Like all regular shapes, a pentagon can be split up into equal triangles, which will be isosceles as two sides, and the corresponding angles are equal.
To find the Area we use the formula for a triangle Area=( X base X height.
We know that the base is 200m since it is the same as one side. But we need to find the height of the triangle to find the Area. We know all the angles in the triangle so we can use trigonometry:
Tan(72/2)= 100/H
Rearranged this formula is:
H=100/Tan46
H=137.64m
Area=( X base X height
Area=0.5 X 200 X 127.64
Area = 13763.82m2
But this is only the Area of one of the triangle so we now have to multiply by 5.
The Area of the Pentagon is=68819.10m2
If I look at the result of this Area, I see that my theory was correct, as the Area of a Pentagon is bigger than the Area of a square. Therefore, I can say that the Area of a regular shape increases as the shape has more sides. I can and will carry on testing this theory with other shapes such as hexagons, heptagons and octagons.
But I have also analysed that the method used to find out the Area of a pentagon is quite lengthy, and it may be appropriate that I find out a formula that will aid me to quickly find out the Area of a regular shape at a later time. Now I will find the Area of a hexagon and see if there is any repetitive method that may help me find a formula.
-Each side of a hexagon= 1000/n or 1000/6= 1662/3m
-Exterior angle= 600 (360/6)
-Interior angle = 1800-600=1200
Tan (300)=83.333/H
H=144.34m
Area=( X base X height
=(X144.34X1662/3
=12028.13m2
Multiply by 6 to find the Area of all the triangles=72168.78m2
From this I can say that the Area does increase with number of sides. The Area has continued rising, as the number of sides has increased, albeit at a slower rate- the difference between the Area of a hexagon and pentagon is not as great as the difference between the Area of a square and triangle.
Now I can see the repetition in some aspects of the working out. I have found that I have been using the Area of a triangle as a foundation, since it has to be used to find the overall Area of a polygon. I have also used the Tan function to find the height of each triangle. So now, with my knowledge so far I can derive a formula to find the Area for an n sided regular polygon.
For the first part of my investigation, I will have to some possible figures for the plot of land, making sure that the perimeter is 1000m.
For my first examination, I will start with quadrilaterals.
Side 1: 100m
Side 2: 400m
Side 1: 200m
Side 2: 300m
Side 1: 250m
Side 2: 250m
Side 1: 450m
Side2: 50m
Side 1: 350m
Side2: 150m
Side 1
Side 2
Area
00 m
400 m
Side*side=
40000 m2
250 m
250 m
Side*side=
62500 m2
300 m
200 m
Side*side=
60000 m2
350 m
50 m
Side*side=
52500 m2
450 m
50 m
Side*side=
22500 m2
The graph and the table indicate that when the width and the height of a regular quadrilateral have the least difference, the Area is at its greatest. This statement is proven by the square. A square has all the sides the same, so the difference is at the smallest since it is at zero. The Area of the square is the greatest, and this leads me to believe that regular shapes- i.e. shapes that have equal sides, have the biggest Area. I will test this prediction out on triangles. I will start with the most regular triangle - the equilateral, and then test a number of isosceles triangles. If my prediction is true then the equilateral triangle will have the biggest Area.
Side 1, 2,3= 333.333
Sin (60)= x/333.33
X=288.66 (height)
Area= ( X base X height
=48112.52m2
Side 2, 3=450m
Base= 100m
150m2 +x2= 350m2
x=316.23m (height)
Area= ( X base X height
= 47494.16m2
Side 2, 3=400m
Base= 200m
00m2 +x2=400m2
x=382.30 (height)
Area=( X base X height
=38729.83m2
Side 2, 3=350m
Base= 300m
50m2+x2=350m2
x=316.23m
Area=( X base X height
=47434.16m2
Side 2, 3=300m
Base=400m
200m2+x2=300m2
x=223.61m
Area=( X base X height
=44721.36m2
Base
Equal sides
Area
00m
450m
22360.68m2
200m
400m
38729.83 m2
300m
350m
47434.16 m2
333(m
333(m
48112.52 m2
400m
300m
44721.36m2
This graph proves my initial theory. The equilateral triangle, which had the least difference between the sides, has the largest Area. The equilateral triangle, which is the most regular of all the triangles since it has equal sides, has the biggest Area. This is similar to the square, which is the most regular of all quadrilaterals. Therefore, I can now use only regular polygons (polygons with equal sides) to gain the biggest Areas. However, I have also made another observation. The three-sided triangle has a smaller Area than the four-sided square. It may be possible that Area increases with the number of sides. Consequently, this is my next hypothesis. To test this new theory, I will have to work out the Area for other regular polygons, with a greater number of sides. I will start off with a pentagon.
- Each side= 200m (1ooo/5)
- All corresponding angles are the same
- Exterior angle = 360o/n or 360/5 = 72o
- Interior angles= 180-72=1080 (angles on a straight line)
Like all regular shapes, a pentagon can be split up into equal triangles, which will be isosceles as two sides, and the corresponding angles are equal.
To find the Area we use the formula for a triangle Area=( X base X height.
We know that the base is 200m since it is the same as one side. But we need to find the height of the triangle to find the Area. We know all the angles in the triangle so we can use trigonometry:
Tan(72/2)= 100/H
Rearranged this formula is:
H=100/Tan46
H=137.64m
Area=( X base X height
Area=0.5 X 200 X 127.64
Area = 13763.82m2
But this is only the Area of one of the triangle so we now have to multiply by 5.
The Area of the Pentagon is=68819.10m2
If I look at the result of this Area, I see that my theory was correct, as the Area of a Pentagon is bigger than the Area of a square. Therefore, I can say that the Area of a regular shape increases as the shape has more sides. I can and will carry on testing this theory with other shapes such as hexagons, heptagons and octagons.
But I have also analysed that the method used to find out the Area of a pentagon is quite lengthy, and it may be appropriate that I find out a formula that will aid me to quickly find out the Area of a regular shape at a later time. Now I will find the Area of a hexagon and see if there is any repetitive method that may help me find a formula.
-Each side of a hexagon= 1000/n or 1000/6= 1662/3m
-Exterior angle= 600 (360/6)
-Interior angle = 1800-600=1200
Tan (300)=83.333/H
H=144.34m
Area=( X base X height
=(X144.34X1662/3
=12028.13m2
Multiply by 6 to find the Area of all the triangles=72168.78m2
From this I can say that the Area does increase with number of sides. The Area has continued rising, as the number of sides has increased, albeit at a slower rate- the difference between the Area of a hexagon and pentagon is not as great as the difference between the Area of a square and triangle.
Now I can see the repetition in some aspects of the working out. I have found that I have been using the Area of a triangle as a foundation, since it has to be used to find the overall Area of a polygon. I have also used the Tan function to find the height of each triangle. So now, with my knowledge so far I can derive a formula to find the Area for an n sided regular polygon.